# Showing A Set is Vector Space

• Check that the set $$\mathbb{R}^\omega$$ of all infinite-tuples of real numbers

$$x=(x_1,x_2,...)$$

is a vector space under component-wise addition and scalar multiplication.

My Attempt. Let $$x,y\in\mathbb{R}^\omega$$. We need to show:

1) $$x+y=(x_1,x_2,...)+(y_1,y_2,...)=(x_1+y_1,x_2+y_2,...)\in\mathbb{R}^\omega$$

2)$$cx=c(x_1,x_2,...)=(cx_1,cx_2,...)\in\mathbb{R}^\omega$$.

How do we know $$1)$$ and $$2)$$ is true? Can you explain, can you show $$1),2)?$$

• 1) and 2) are true because you define addition and scalar multiplication like that. Indeed "compenent-wise addition and scalar multiplication" mean exactly that. Now, what you have to prove is not the definition, but that those definitions satisfy vector space's axioms. Do you know those axioms (i.e. "A vector space is...")? – Crostul Jul 28 '19 at 12:47
• A nice list of the axioms is given on the wiki page. – Omnomnomnom Jul 28 '19 at 14:14
• @Crostul Yes, I know – James Ensor Jul 30 '19 at 14:44
• @Omnomnomnom Thanks... – James Ensor Jul 30 '19 at 14:44
• I posted an answer, can you check? Is my answer enough? – James Ensor Jul 30 '19 at 14:56

At first we have to explain why $$(x_1 + y_1, x_2 + y_2, \dots)$$ and $$(cx_1, cx_2, \dots)$$ are elements of $$\mathbb R^\omega$$. Note that this is true because the sum of two real numbers is again a real number and the product of two real numbers is again a real number.

So what we now have to do is to verify the eight vector space axioms (see e. g. the wiki page). I will give you here the proof of two of them and let you the proof of the remaining ones as an exercise.

Commutativity of addition: Let $$x = (x_1, x_2, \dots), y = (y_1, y_2, \dots)$$ as above. Then

$$x + y = (x_1 + y_1, x_2 + y_2, \dots)\ {\color{red} = }\ (y_1 + x_1, y_2 + x_2, \dots) = y + x.$$

Here, the crucial point is to explain why the red marked equality holds. This is the case because of the commutativity of the addition in $$\mathbb R$$: We have $$a + b = b + a$$ for all $$a, b \in \mathbb R$$. (Remark: You will need the properties of $$\mathbb R$$ for showing that $$\mathbb R^\omega$$ is a vector space, everything follows from this.)

Existence of a zero vector: We have to show that there is a $$0_{\mathbb R^\omega}$$ such that $$x + 0_{\mathbb R^\omega} = x = 0_{\mathbb R^\omega} + x$$ for all $$x \in \mathbb R^\omega$$. Note that I write $$0_{\mathbb R^\omega}$$ more because of pedagogical reasons, this distinguishes the zero vector from $$0 \in \mathbb R$$. So take

$$0_{\mathbb R^\omega} = (0, 0, \dots) \in \mathbb R^\omega.$$

Then

\begin{align} x + 0_{\mathbb R^\omega} &= (x_1 + 0, x_2 + 0, \dots ) = (x_1, x_2, \dots) = x , \\ 0_{\mathbb R^\omega} + x &= (0 + x_1, 0 + x_2, \dots ) = (x_1, x_2, \dots) = x. \end{align}

Here we use again a fact of the real numbers: The zero element is there $$0$$, so we have $$x_1 + 0 = 0 + x_1 = x_1$$ for every real $$x_1$$. We have established the existence of a zero vector, you now have to prove the remaining six axioms.

• Thanks for answer and comments – James Ensor Aug 1 '19 at 10:11

Let $$x,y\in\mathbb{R}^\omega$$. Since we know 1) $$x+y=(x_1,x_2,...)+(y_1,y_2,...)=(x_1+y_1,x_2+y_2,...)\in\mathbb{R}^\omega$$

2)$$cx=c(x_1,x_2,...)=(cx_1,cx_2,...)\in\mathbb{R}^\omega$$,

then $$\mathbb{R}^\omega$$ is a vector space.

• In order to let this be a valid answer you should provide a larger vector space $X$ such that $\mathbb{R}^\omega \subset X$, i. e., $\mathbb{R}^\omega$ is a subspace of $X$. Otherwise you have to show all vector space axioms. – Jan Jul 30 '19 at 15:00
• Okey, thanks... – James Ensor Jul 30 '19 at 15:04